Exercise 6.4.2

Proof. First, we show for normal operator $A$, $\left|\right|Ax\left|\right|=\left|\right|A^{\text{∗}}x\left|\right|$. Note that $A{A}^{\text{∗}}=A^{\text{∗}}A$, then

Thus $\left|\right|Ax\left|\right|=\left|\right|A^{\text{∗}}x\left|\right|$.

Suppose $v$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$. Note that $A-\mathit{\lambda I}$ is also normal (see Problem 2.2). Thus we have $\left|\right|(A-\mathit{\lambda I})v\left|\right|=\left|\right|{(A-\mathit{\lambda I})}^{\text{∗}}v\left|\right|=\left|\right|(A^{\text{∗}}-\overline{\lambda }I)v\left|\right|=0$, i.e., $A^{\text{∗}}v=\overline{\lambda }v$. So $A^{\text{∗}}Av=A^{\text{∗}}(\lambda v)=\lambda \overline{\lambda }v=|\lambda {\text{|}}^{2}v$. $|\lambda {\text{|}}^2$ is an eigenvalue of $A^{\text{∗}}A$, then $\left|\lambda \right|$ is a singular value of $A$. □